Hallo,
can you help me writing a generic xslt transformation (useable with xsql from oracle)? The problem is how to get the escaping characters ....
=== INPUT-File in.xml
<?xml version = '1.0'?>
<person><who>scott</who></person>
=== Needed OUTPUT-File out.xml
=== Note: the format must be exactly this to be
=== processable by xsql using xsql-insert
<?xml version = '1.0'?>
<ROWSET><ROW>
<NAME><person><who>scott</who></person></NAME> </ROW></ROWSET>
=== My xslt-File p.xsl, that misses the important fact
=== of replacing < with <
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0"> <xsl
utput method="xml" omit-xml-declaration="yes"/> <xsl:template match="@*|node"> <xsl:element name="ROWSET"> <xsl:element name="ROW"> <xsl:element name="NAME"> <xsl:copy-of select="."/> </xsl:element> </xsl:element> </xsl:element> </xsl:template> </xsl:stylesheet>
=== my output, that is not the expected
oraxsl in.xml p.xsl
<ROWSET><ROW><NAME><person><who>scott</who></person></NAME></ROW></ROWSET>
Any ideas?
Thank you if you find the time for helping,
Martin
can you help me writing a generic xslt transformation (useable with xsql from oracle)? The problem is how to get the escaping characters ....
=== INPUT-File in.xml
<?xml version = '1.0'?>
<person><who>scott</who></person>
=== Needed OUTPUT-File out.xml
=== Note: the format must be exactly this to be
=== processable by xsql using xsql-insert
<?xml version = '1.0'?>
<ROWSET><ROW>
<NAME><person><who>scott</who></person></NAME> </ROW></ROWSET>
=== My xslt-File p.xsl, that misses the important fact
=== of replacing < with <
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0"> <xsl
=== my output, that is not the expected
oraxsl in.xml p.xsl
<ROWSET><ROW><NAME><person><who>scott</who></person></NAME></ROW></ROWSET>
Any ideas?
Thank you if you find the time for helping,
Martin